1) Basic Definitions 2) Divisibility Rules 3) No. of Factors 4) No. of Ways 5) Sum of Factors 6) Units Digit 7) Types of Number Systems 8) Highest Power
Concept 1:
Basic Definitions:
In this, we will learn about all definitions in number systems, such as natural numbers, whole numbers, integers, rational numbers, irrational numbers, real numbers, imaginary numbers, complex numbers, prime numbers, composite numbers, twin prime numbers, co - prime numbers, prime triplet etc.
Concept 2:
Divisibility Rules:
In this concept, we will learn about all divisibility rules from 1 to 20 and along with those divisibility rule we will also learn about some other divisibility rules which are very important for examination point of view.
Eg: Check whether the number 3794 is divisible by 7 or not.
Sol: 3794
A number is divisible by 7, if the difference between number of tens and twice the unit’s place
is divisible by 7.
No. of tens = 379, twice the units place = 2 × 4 = 8,
Difference = 379 – 8 = 371
Again, follow the same rule for 371
No. of tens = 37, twice the unit’s place = 2 × 1 = 2
Difference = 37 – 2 = 35
Since 35 is divisible by 7, hence 3794 is divisible by 7.
Eg: Check whether the number 26543 is divisible by 11 or not.
Sol: 26543
A number is divisible by 11, if the difference between the sum of odd place digits and the sum of even place digits is either zero (or) multiple of 11.
Sum of odd place digits = 2 + 5 + 3 = 10
Sum of even place digits = 6 + 4 = 10
Difference = 10 – 10 = 0
Hence, the number 26543 is divisible by 11.
Concept 3:
No. of Factors:
In this concept, we will learn about how to calculate number of factors of a given number. Knowing how to calculate number of factors is must, because to calculate number of ways, candidate needs to calculate number of factors first.
Let us consider a given number N, write that number interms of prime factors.
N = P1a × P2b × P3c × ……….
Where P1, P2, P3, ….. are prime numbers.
a, b, c …… are natural numbers.
No. of factors = (a + 1)(b + 1)(c + 1) …….
Eg: Find the number of factors for the number 196.
Sol: 196 = 22 × 72
No. of factors = (2 + 1)(2 + 1) = 3 × 3 = 9 factors
Concept 4:
No. of Ways:
In this concept, we will learn about how to calculate number of ways of a given number. Number of ways are always depending on number of factors.
There are two cases to find the number of ways.
Case 1: If number of factors are even, then
Number of ways = No. of Factors / 2
Eg: Find the number of ways for the number 24.
Sol: 24 factors are 1, 2, 3, 4, 6, 8, 12, 24
No. of factors = 8 (even number)
No. of ways = 8 / 2 = 4 ways.
Case 2: If number of factors are odd, then again we are having 2 conditions
a) as a product of 2 factors
No. of ways = (No. of factors + 1) / 2
Eg: Find the number of ways for the number 36.
Sol: 36 factors are 1, 2, 3, 4, 6, 9, 12, 18, 36
No. of factors = 9 (odd number)
No. of ways = (9 + 1) / 2 = 5 ways.
b) as a product of 2 different factors
No. of ways = (No. of factors - 1) / 2
Eg: Find the number of ways for the number 36.
Sol: 36 factors are 1, 2, 3, 4, 6, 9, 12, 18, 36
No. of factors = 9 (odd number)
No. of ways = (9 - 1) / 2 = 4 ways.
Concept 5:
Sum of Factors:
In this concept, we will learn about how to calculate sum of factors of a given number.
Let us consider a given number N, write that number interms of prime factors.
N = P1a × P2b × P3c × ……….
Where P1, P2, P3, ….. are prime numbers.
a, b, c …… are natural numbers.
Sum of factors = [P1a + 1 - 1] / [P1 - 1] × [P2b + 1 - 1] / [P2 - 1] × [P3c + 1 - 1] / [P3 - 1] ....
Eg: Find the sum of factors for the number 12.
Sol: 12 = 22 × 31
Sum of factors = [22 + 1 - 1] / [2 - 1] × [31 + 1 - 1] / [3 - 1] = 7 × 8 / 2 = 28
Concept 6:
Units Digit:
In this concept, we will learn about how to calculate units place digit of any exponential number with higher order. This concept is very helpful in examination point of view, when we are solving the problems through option verification method.
Eg: Find the units place digit of (562)327.
Sol: Unit’s place digit in (562)327 = 2Remainder
327 / 4 => Remainder = 3
Unit’s place digit in (562)327 = 23 = 8.
Concept 7:
Types of Number Systems:
In this concept, we will learn about the various types of number systems. Also we will learn how to convert one type of number system to another type of number system. Basically, we are having 4 types of number systems.
1. Binary Number System
Binary number system consisting of only 2 digits, those are 0 and 1. It is denoted by base 2.
2. Octal Number System
Octal number system consisting of 8 digits from 0 to 7. It is denoted by base 8.
3. Decimal Number System
Decimal number system consisting of 10 digits from 0 to 9. It is denoted by base 10. In general we are using decimal number system.
4. Hexa decimal Number System
Hexa decimal number system consisting of 16 digits from 0–9, A–F. It is denoted by base 16. In this A = 10, B = 11, C = 12, D = 13, E = 14 and F = 15.
Concept 8:
Highest Power of Prime number in N!:
To solve the problems in this concept, first we have to check whether the given number is prime (or) not. If it is not a prime number, then convert that number into prime factors. Here ‘N’ is any natural number.
Highest power of a prime number in N! = Sum of all quotients / Power of Prime number
Eg: Find the highest power of 2 in 10!.
Sol: Here ‘2’ is a prime number and
10! = 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10
Write every possible number interms of 2
10! = 1 × 21 × 3 × 22 × 5 × (21 × 31) × 7 × 23 × 9 × (21 × 51)
10! = 28 × 32 × 52 × 7 × 9
Highest power of 2 in 10! is 8.
Important Formulae:
1. (a + b)² = a² + 2ab + b²
2. (a – b)² = a² – 2ab + b²
3. (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
4. (a + b)³ = a³ + b³ + 3ab (a + b)
5. (a – b)3 = a3 – b3 – 3ab (a – b)
6. (a + b + c)3 = a³ + b³ + c³ + 3 (a + b)(b + c)(c + a)
7. a² – b² = (a + b)(a – b)
8. (a + b)² – (a – b)² = 4ab
9. a³ + b³ = (a + b)(a² – ab + b²)
10. a³ – b³ = (a – b)(a² + ab + b²)
11. a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
How many numbers up to 600 are divisible by both 2 and 9?
a) 27 b) 28 c) 31 d) 33
Solution:
If a number is divisible by both 2 and 9, then that number must be divisible by 18 (LCM of 2, 9)
⸫ 33 numbers are divisible by both 2 and 9 i.e; 18 upto 600.
Answer: d
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